158. Read N Characters Given Read4 II - Call multiple times
难度: 困难
刷题内容
原题连接
- https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times
内容描述
The API: int read4(char *buf) reads 4 characters at a time from a file.
The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.
By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file.
Note:
The read function may be called multiple times.
Example 1:
Given buf = "abc"
read("abc", 1) // returns "a"
read("abc", 2); // returns "bc"
read("abc", 1); // returns ""
Example 2:
Given buf = "abc"
read("abc", 4) // returns "abc"
read("abc", 1); // returns ""
解题方案
思路 1
这个题目的描述就是一坨屎💩，真的，sb。
我来总结一下，跟第157题不一样的地方就是，157是就读一次，158是可以读好几次 例如： 文件是‘abcdefg’ - 157题就读一次，给一个n就行了。n给1那buf就是‘a’, n给2那buf就是‘ab’ - 但是158不一样，可以多次read，比如第一次n给1，那buf是‘a’，再read一次，n给2，那'a'已经读过了，所以现在buf是'bc'了， 如果再来个n=3的话，buf就是‘def’,
总之就是一个test case 中read函数可以调用一次和调用多次的区别
class Solution(object):
head, tail, buffer = 0, 0, [''] * 4 ## 定义全局变量
def read(self, buf, n):
"""
:type buf: Destination buffer (List[str])
:type n: Maximum number of characters to read (int)
:rtype: The number of characters read (int)
"""
i = 0
while i < n:
if self.head == self.tail: ## read4 的缓存区为空的时候
self.head = 0
self.tail = read4(self.buffer) ## 开始进缓存区
if self.tail == 0:
break
while i < n and self.head < self.tail:
buf[i] = self.buffer[self.head] ## 读出缓存区的变量
i += 1
self.head += 1
return i
